Let $x_1, \ldots, x_n \in \mathbb{R}^d$ denote $n$ points in $d$-dimensional Euclidean space, and $w_1, \ldots, w_n \in \mathbb{R}_{\geq 0}$ any non-negative weights.
$\arg\min_{\mu \in \mathbb{R}^d} \sum_{i=1}^n w_i | x_i-\mu| = median\{x_1, \ldots, x_n \}?$
I understand why the median of $\{x_1, \ldots, x_n \}$ minimizes the function without the weights. But I am not sure if the median is what minimizes the function with the weights, should I use Lagrange multiplier to solve this?
On
The problem you have stated is the weighted L1 median (sometimes called the weighted geometric median) of your data. There are many references on the topic: https://www.google.com/search?q=l1+median
A concise reference along with an algorithm is: http://www.stat.rutgers.edu/home/cunhui/papers/39.pdf
However, note that the Vardi-Zhang process needs to be adjusted for data having repeated values. If you scan the literature, there are some references that describe this adjustment.
What minimizes your weighted is again the median !!! But not the median of $x_1,x_2,... $.
A value that minimizes your sum will be a number $\mu $ such that
$$ \sum_{i\in \{1,2,...,n\}, x_i>\mu} w_i < \frac {\sum_{i=1}^n w_i}{2} $$
$$ \sum_{ i\in \{1,2,...,n\},\,\, x_i<\mu} w_i < \frac {\sum_{i=1}^n w_i}{2} $$
Note that this $\mu $ will also a median. It is a median of the random variable defined on the sample space ${1,2,...,n} $ that sends $i $ to $x_i $. In this probability space, the probability of occurrence of $i $ is $\frac {w_i} {w_1+...+w_n} $
Opps! ! I just noted that your $x_i $ can be vectors, therefore my answer only works when $d=1$